Possible Duplicate:
Invariants and base change
Suppose I have a commutative ring $R$, a free module $M$ of finite rank, and a finitely generated group $G$ acting on $M$ via $R$-linear endomorphisms.
Is there a nice condition on the action of $G$ on $M$ that would imply that taking invariants always commutes with base change, i.e. $$(M \otimes S)^{G} = M^G \otimes S$$ for any $R$-algebra $S$?
